1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Problems with mechanics

  1. Jan 17, 2009 #1

    I'm having trouble with my mechanics course. I'm having trouble figuring out what the question is asking me with all this polar coordinare stufff, and how to go about the question. Here are some of the questions.

    1. A particle starts on the x axis at (b, 0) at time t = 0, and moves at constant speed v in a straight line in the +y direction. Derive explicit expressions (in terms of b, v, and t) for the polar coordinates r and f as functions of time, and for the time derivatives f fr&, &, &r&, and && . Finally, calculate the radial and azimuthal acceleration components ar and af.

    My general approach is to wrtite the pythagorean therorem for distace after time t to get the radious, and then use trig to get an expression for the angle. But, then I don't really know what to do get the derivatives.

    There are other questions but I'm not sure how to start them, and so don't have an attempted solution.


  2. jcsd
  3. Jan 17, 2009 #2
    Here's a place to start: the particle's position as a function of time in Cartesian co-ordinates is just

    (1) [tex]\stackrel{\rightarrow}{r} =b\stackrel{\rightarrow}{x} + vt\stackrel{\rightarrow}{y}[/tex]

    What is the velocity?

    (2)[tex] \frac{d}{dt}\stackrel{\rightarrow}{r} = \frac{d}{dt}(b\stackrel{\rightarrow}{x}) + \frac{d}{dt}(vt\stackrel{\rightarrow}{y}}) = v\stackrel{\rightarrow}{y} [/tex]

    One can write the Cartesian unit vectors in terms of the polar unit vectors. Go for it! :tongue2:

    If a particle is moving with a constant speed in a straight line, what is its acceleration? And, if that is its acceleration, what must its components be?
  4. Jan 17, 2009 #3
    Here's another way to approach it:

    r = (b2 + (vt)2)1/2 <r> where <r> is the unit vector in the radial direction. θ = arctan(vt/b). The position vector r is a function of time that can be expressed as

    r = r(t) <r>

    Then dr/dt = dr/dt <r> + r(t) d<r>/dθ dθ/dt. And d<r>/dt = <θ>, the unit vector in the theta direction, and dθ/dt is the azimuthal velocity. This gives

    v = dr/dt = dr/dt<r> + r(t)dθ/dt <θ>

    Find a = dv/dt and use d<θ> /dθ = -<r>.
  5. Jan 18, 2009 #4
    Thanks to both of you for replyingm, although I'm doing it closer to Chris's way.

    I have the first part done exactlty the same. Where you have an r in bold(not wwhere you have an r hat in bold) is that where I need to take the derivative in terms of t? Likewise with theta?

    Also, what is r not in bold format? Or is ithe other way round? Our prof showed us this in class, but didn't do a great job of it, plus he uses different notation.

    Lastly what is dr/dt? Is it the dervative of r(t), where r(t) is bold r, minus the r hat?
    Last edited: Jan 18, 2009
  6. Jan 18, 2009 #5
    Yes, take the derivative of r(t) <r> with respect to time using the chain rule. The <r> is the unit vector in the r direction. Recall, a vector has a magnitude, in this case r(t), and a direction, <r>. The <r> is called a unit vector because it only gives a direction since the magnitued is equal to 1 (unity). The difference between the polar unit vectors and the cartesian unit vectors <x> and <y> are the polar unit vectors change direction as r changes whereas the cartesian unit vectors remain fixed. Therefore, the change in the polar unit vector directions must be evaluated with respect to time.

    There's no need to take the derivative of the theta function in terms of arctan because the unit vector <θ> will show up when taking the first derivative as shown previously.

    The dr/dt is simply the derivative of r(t) with respect to time and this is a magnitude, not a vector. The -<r> is a unit vector in the negative <r> direction.
  7. Jan 18, 2009 #6
  8. Jan 18, 2009 #7
    Sorry if I confused you. The dθ/dt term is the azimuthal velocity or angular velocity magnitude and is usually expressed as ω. When the acceleration is found the dω/dt term is usually expressed as α (alpha), the azimuthal or angular acceleration. So, if you desire to find dθ/dt as a function of time, yes, take the derivative of the θ = arctan expression with respect to time.
  9. Jan 18, 2009 #8
    So, d(theta)/dt = omega, and d^2(theta/dt^2 + alpha, so we never actually have to take the derivative of theat at all, and becuase we want azimuthal accelaration we leave thse terms as omega and alpha.

    Won't we need to calculate omgega to find out what alpha is? Or is it just asking for an expression fot the ra and aa.

    Radial accelaration is the 2nd derivative of the r equation I assume.
  10. Jan 18, 2009 #9
    I dont really understand what a(r) and a(theta) are.
  11. Jan 18, 2009 #10
    Take the the time derivative of θ = arctan(vt/b). This will give ω as a function of time. And dω/dt = α and will be a funtion of time also. Yes, one component of the radial acceleration is the second time derivative of r(t). You will have four terms for the acceleation vector; it comes from using the chain rule for the v = dr/dt expression; two terms will be in the radial direction and two will be in the azimuthal direction.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook